We discuss the important class of continuous functions of one variable. We teach you how to prove that a function is continuous. We also provide several examples and exercises with detailed answers to illustrate our course.

To fully understand the content of this page, some tools on the limit of functions are necessary.

Generalities on continuous functions

Throughout this paragraph,  $\mathscr{D}$ is a subset of $\mathbb{R}$ and $f:\mathscr{D}\to \mathbb{R}$ is a function. The set $\mathscr{D}$ is called the domain of $f$.

Definition: We say that $f$ is continuous at point $a\in D$ if $f$ admit a limit $f(a)$ at the point $a$. That is, for any $\varepsilon>0,$ there exists $\delta>0$, such that for any $x\in D,$ $|x-a|<\delta$ implies that $|f(x)-f(a)|<\varepsilon$.

We say that $f$ is continuous on $\mathscr{D}$ if $f$ is continuous on each point of $\mathscr{D}$.

If $f$ and $g$ are continuous functions, the sum, the product, the composition, and the quotient of these functions are also continuous.

Classical examples: All polynomial functions are continuous on $\mathscr{D}=\mathbb{R}.$ The square root function $x\mapsto \sqrt{x}$ is continuous on $\mathscr{D}=[0,+\infty)$. The trigonometric function $x\mapsto \sin(x)$ and $x\mapsto \cos(x)$ are continuous on $\mathbb{R}$. The logarithmic function $x\mapsto \ln(x)$ is continuous on $\mathscr{D}=(0,+\infty)$. The exponential function $x\mapsto e^x$ is continuous on $\mathscr{D}=\mathbb{R}$.

The other functions are usually the sum, product, composition, inverse, and quotient of the classical functions above.

Continuity using sequences: the function $f$ is continuous at $a\in \mathscr{D}$ if and only if for any sequence $(u_n)_n$ of $\mathscr{D}$ converging to $0,$ the image sequence $( f (u_n))_n$ converges to $f(a)$.

This result is very useful if we want to prove that the function $f$ is not continuous at a point $a$. To do this, it suffices to find two sequences both converging towards $a,$ while their images converge towards different points.

Theorem: If $f:[a,b]\to\mathbb{R}$ is continuous then $f$ is bounded on $[a,b]$, this is there exists a real number $M>0$ such that $|f(x)|\le M$ for any $x\in [a,b]$. Moreover, there exists $\alpha,\beta\in [a,b]$ such that \begin{align*} f(\alpha)=\sup_{x\in [a,b]}f(x),\qquad f(\beta)=\inf_{x\in [a,b]}f(x).\end{align*}

A selection of exercises on continuity of functions

Exercise: Determine if the following functions are continuous or not: \begin{align*} f(x)=\begin{cases}\displaystyle\frac{\sin(2x)-\sin(5x)}{\sin(3x)},& x\neq 0,\cr -1,& x=0.\end{cases}\qquad g(x)=\begin{cases} x^x,& x>0,\cr 2,&x=0.\end{cases} \end{align*}

Solution: 1) The function $f$ is continuous on $\mathbb{R}\backslash\{0\}$ as the sum and the quotient of continuous functions. Now let us verify the continuity of $f$ at $0$. For that purpose, we shall calculate the limit of $f$ as $0$ and check if it is equal to $f(0)$. Before doing so we recall the following standard result \begin{align*} \lim_{x\to 0,\;x\neq 0} \frac{\sin(x)}{x}=1. \end{align*} Now for any $x\neq 0,$ we have \begin{align*} f(x)&=\frac{x}{\sin(3x)}\times \left(\frac{\sin(2x)}{x}-\frac{\sin(5x)}{x}\right)\cr &= \frac{1}{3}\frac{3x}{\sin(3x)}\times \left(2\frac{\sin(2x)}{2x}-5\frac{\sin(5x)}{5x}\right). \end{align*} Hence \begin{align*} \lim_{x\to 0,\;x\neq 0}f(x)=\frac{1}{3}(2-5)=-1=f(0). \end{align*} It follows that $f$ is continuous on $\mathbb{R}$.

2) For any $x>0$ we can write \begin{align*} g(x)=e^{\ln(x^x)}=e^{x \ln(x)}. \end{align*} This shows that $g$ is continuous on $(0,\infty)$. On the other hand we now that \begin{align*} \lim_{x\to 0^+}x\ln(x)=0. \end{align*} This implies that \begin{align*} \lim_{x\to 0^+} g(x)=1\neq g(0). \end{align*} Then $g$ is not continuous at $0$.

Exercise: Consider the function \begin{align*} f(t)=\begin{cases}\frac{\ln(1+2t^2)}{\sin^2(t)} ,& t > 0,\cr 0,& x=2. \end{cases} \end{align*} Prove that $f$ is continuous on $[0,+\infty)$.

Solution: The function $f$ is continuous on $(0,+\infty)$, as the quotient of continuous functions. We now show what happens if $t$ is very closed to $0$. For $t\neq 0,$ we have \begin{align*} f(t)=2\frac{\ln(1+2t^2)}{2t^2}\left(\frac{t}{\sin(t)}\right)^2 \end{align*} As \begin{align*} \lim_{t\to 0,t\neq 0} \frac{\ln(1+2t^2)}{2t^2}=1, \end{align*} then \begin{align*} \lim_{t\to 0,t\neq 0}f(x)=2=f(0). \end{align*} This shows that $f$ is also continuous at $0$.

Exercise: Let $f$ be defined by \begin{align*} f(x)=\begin{cases} 1,& x\in \mathbb{Q},\cr 0,& x\notin \mathbb{Q}.\end{cases} \end{align*} Prove that $f$ is not continuous at any point of $\mathbb{R}$.

Solution: We recall that $g:A\subset \mathbb{R}\to \mathbb{R}$ is continuous at $x\in A$ if and only if for any sequence $(u_n)_n\subset A$ such that $u_n\to x$ as $n\to\infty,$ we have $g(u_n)\to g(x)$.

Using this result, to prove that $f$ is not continuous at $x\in \mathbb{R}$ it suffices to find two sequences $(u_n)_n$ and $(v_n)_n$ both converging to $x$ as $n\to \infty$ but the limits of $f(u_n)$ and $f(v_n)$ as $n\to\infty$ are different. In fact, let $x\in \mathbb{R}$. By density of $\mathbb{Q}$ in $\mathbb{R}$ there exists $(u_n)_n\subset \mathbb{Q}$ such that $u_n\to x$ as $n\to \infty$. By definition of $f$ we have $f(u_n)=1$, so the limit of $f(u_n)$ at infinity is $1$. On the other hand, by the density of $\mathbb{R}\backslash\mathbb{Q}$ in $\mathbb{R}$ there exists $(v_n)_n\subset \mathbb{R}\backslash\mathbb{Q}$ such that $v_n\to x$ as $n\to infty$. By definition of $f,$ we have $f(v_n)=0,$ so the limit of $f(v_n)$ at infinity is $0$. We then conclude that $f$ is not continuous at $x$.

In this article, we propose an exercise that describes the optimal method of the step gradient for coercive functions on spaces of finite dimension. These specific problems are used in optimization theory and some applications in finance. We mention also we can use the convex functions to study the optimality of some models.

A problem with the optimal method of the step gradient

Problem: Let $f:\mathbb{R}^n\to \mathbb{R}$ be a function of class $C^1$. Suppose that there exists a real number $\alpha>0$ such that \begin{align*} \forall (u,v)\in \mathbb{R}^n\times \mathbb{R}^n,\qquad \langle \nabla f(v)-\nabla f(u),v-u\rangle \;\ge\alpha |v-u|^2. \end{align*}

• Prove that \begin{align*} \forall (u,v)\in \mathbb{R}^n\times \mathbb{R}^n,\qquad f(v)\ge f(u)+ \langle\nabla f(u),v-u\rangle+\frac{\alpha}{2} \|v-u\|^2. \end{align*}
• Deduce that $f$ admits a global minimum on $\mathbb{R}^n$, reached in a unique point that will be denoted by $a$.
• Let $u\in \mathbb{R}^n\backslash\{a\}$. Consider the function \begin{align*}\varphi_u:\mathbb{R}\to \mathbb{R},\quad t\mapsto f(u+t\nabla f(u)). \end{align*} Prove that $\varphi_u$ admits a global minimum on $\mathbb{R},$ reached in a unique point. This allows us to define a sequence $(u_k)_{k\ge 0}$ in $\mathbb{R}^n$ such that: $u_0\in \mathbb{R}^n$ ; if $k_k=a,$ we select $u_{k+1}=a,$ if not, let $t_k$ be the unique real number such that $\varphi_{u_k}(t_k)=\min(\varphi_{u_k})$. We then set \begin{align*} u_{k+1}=u_k+t_k \nabla f(u_k). \end{align*}
• Verify that for any $k\in\mathbb{N},$ $\nabla f(u_k)\bot \nabla f(u_{k+1})$.
• Prove that the series $(u_{k+1}-u_k)_{k\ge 0}$; $(\nabla f(u_{k+1})-\nabla f(u_k))_{k\ge 0}$ and $(\nabla f(u_k))_{k\ge 0}$ the suites tend towards $0$. Deduce that $u_k\to a$ as $k\to\infty$.

Riemann integral exercises with detailed answers are offered on this page. These kinds of integrals of bounded function in bounded intervals, almost everywhere continuous. Thus, all continuous or monotone functions are Riemann integrable.

Lower integral, upper integral, and proper integral 

In this section, we give a concise definition of Riemann integrals. To this end, we assume that a real function is defined and bounded on the interval $[a,b]$, with $a,b\in\mathbb{R}$ are such that $a<b$.

For a subdivision $\sigma:=\{x_0,\cdots.x_n\}$ of the interval $[a,b]$, i.e. $a=x_0<x_1<\cdots<x_n=b$, we define the lower sum of $f$ on $[a,b]$, by $$ L_n(f,\sigma)=\sum_{i=1}^n m_i(f,\sigma)(x_i-x_{i-1}),$$ where $m_i(f,\sigma)$ is the infinimum of $f$ on the interval $[x_{i-1},x_i]$.

By definition, the lower integral of f over the interval $[a,b]$ is $$ \underline{\int^b_a}f(x)ds=\sup_{\sigma} L_n(f,\sigma)=\lim_{n\to\infty}U_n(f,\sigma).$$ Similarly, we define the upper sum of $f$ over the interval $[a,b]$ by $$ U_n(f,\sigma)=\sum_{i=1}^n M_i(f,\sigma)(x_i-x_{i-1}),$$ where $M_i(f,\sigma)$ is the supremum of $f$ on the interval $[x_{i-1},x_i]$.

The upper integral of f over the interval $[a,b]$ is $$ \overline{\int^b_a}f(x)ds=\inf_{\sigma} U_n(f,\sigma)=\lim_{n\to\infty}U_n(f,\sigma).$$ Definition: A bounded function $f:[a,b]\to\mathbb{R}$ is said to be Riemann integrable on $[a,b]$ if its lower and upper integrals coincide. In this case,  the integral of $f$ over $[a,b]$ is given by $$\int^b_a f(x)dx=\overline{\int^b_a}f(x)ds=\underline{\int^b_a}f(x)ds.$$ There exists another equivalent condition for Riemann integrals. Let’s give a summary of it. We take $c_\in [x_{i-1},x_i]$ for $i=1,\cdots,n$ and de define the Riemann sum by $$S_n(f,\sigma):=\sum_{i=1}^n f(c_i)(x_{i}-x_{i-1}).$$ Observe that $L_n(f,\sigma)\le S_n(f,\sigma)\le U_n(f,\sigma)$. Thus if $f$ is Riemann integrable then $S_n(f,\sigma)$ has a limit as $n\to\infty,$ and in this case we have $$ \int^b_a f(x)dx=\lim_{n\to\infty} \sum_{i=1}^n f(c_i)(x_{i}-x_{i-1}).$$

A particular cas: Let $f:[0,1]\to \mathbb{R}$ be Riemann integral and take the uniform subdivision $\sigma:=\{\frac{i}{n}: i=0,\cdots,n\}$, and $c_i=\frac{i}{n}$, Then $$ \int^1_0 f(x)dx=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f\left(\frac{i}{n}\right).$$

Riemann integral exercises

We start our selection of Riemann integral exercises by providing an example of a non-integrable bounded function in the sense of Riemann

Exercise:  Prove that the following function $$g(x)=\cos^2(x)1_{\mathbb{Q}}(x),\quad x\in [0,\pi/2],$$ is not Riemann integrable.

Exercise: Let $f:[0,1]\to \mathbb{R}$ be a continuous function. For any $n\in\mathbb{N},$ we set \begin{align*} u_n=n\int^1_0 t^n f(t)dt. \end{align*}

1. Let $\varepsilon>0$. Prove that there exists $a\in (0,1)$ such that $|f(t)|\le \frac{\varepsilon}{2}$ whenever $t\in [a,1]$.
2. We denote $M=\sup{|f(t)|:t\in [0,1]}$, a such $M\in \mathbb{R}^+$ exists because $f$ is continuous on the compact $[0,1]$. Prove that \begin{align*} |u_n|\le M a^n+\frac{\varepsilon}{2},\qquad \forall n\in\mathbb{N}. \end{align*}
3. Deduce that $\lim_{n\to+\infty}u_n=0$.
4. Deduce that, in general, we have \begin{align*} \lim_{n\to+\infty}u_n=f(1). \end{align*}

Solution: The first 3 question: By definition of left continuity at point $1,$ for all $\varepsilon>0,$ there exists $0 < \delta < 1$ such that $t\in [1-\delta,1]$ we have $|f(t)-f(1)|=|f(t)| \le \frac{\varepsilon}{2}$. It suffices to put $a=1-\delta\in (0,1)$.

We can write \begin{align*} |u_n|&\le n \int^1_0 t^n |f(t)|dt\cr &\le n \int^a_0 t^n |f(t)|dt+n \int^1_a t^n |f(t)|dt\cr & \le n M \int^a_0 t^n dt+ n \frac{\varepsilon}{2} \int^1_a t^n dt\cr & \le n M \left[\frac{t^{n+1}}{n+1}\right]^a_0+ n \frac{\varepsilon}{2} \left[\frac{t^{n+1}}{n+1}\right]^1_a \cr &\le M \frac{n}{n+1} a^{n+1}+\frac{\varepsilon}{2} \frac{n}{n+1} (1-a^{n+1}). \end{align*} As $a\in (0,1),$ then $a^{n+1}\le a^n$ and $0 < 1-a^{n+1} < 1$. On the other hand, $\frac{n}{n+1}le 1$. This implies that \begin{align*} |u_n|\le M a^n+\frac{\varepsilon}{2},\qquad \forall n\in\mathbb{N}. \end{align*}

Since $a\in (0,1)$ then the geometric sequence $(a^n)_n$ satisfies $a^n\to 0$ as $n\to+\infty$. Then there exists $N\in \mathbb{N}$ such that for any $n\ge N,$ we have $0 < a^n < \frac{\varepsilon}{2M}$. Hence for $n\ge N,$ we have \begin{align*} |u_n|\le M \frac{\varepsilon}{2M}+\frac{\varepsilon}{2}=\varepsilon. \end{align*} Thus $u_n\to 0$ as $n\to +\infty$.

4) In this question we work with a general continuous function $f:[0,1]\to \mathbb{R}$. We set $g(t)=f(t)-f(1)$. The function $g$ is continuous on $[0,1]$ and satisfies $g(1)=0$. According to the first question we have \begin{align*} \lim_{n\to+\infty} n\int^1_0 t^n g(t)dt=0. \end{align*} In addition, observe that \begin{align*} u_n&=n\int^1_0 t^n g(t)dt+n f(1) \int^1_0 t^n dt \cr &= n\int^1_0 t^n g(t)dt+\frac{n}{n+1}f(1) \end{align*} Hence $u_n\to f(1)$ as $n\to +\infty$.

Exercise: Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function and let $\omega\in\mathbb{R}^\ast$. Define the following function \begin{align*} \varphi(x)=\frac{1}{\omega}\int^x_0 \sin\left(\omega (x-t)\right)f(t),dt,\qquad x\in\mathbb{R}. \end{align*}

• Prove that there exist two functions $\Phi,\Psi:\mathbb{R}\to \mathbb{R}$ such that for any $x\in\mathbb{R},$ \begin{align*} \varphi(x)=\frac{\sin(\omega x)}{\omega} \Phi(x)- \frac{\cos(\omega x)}{\omega}\Psi(x). \end{align*}
• Show that $\varphi$ is twice differentiable on $\mathbb{R}$ and that \begin{align*} \varphi”+\omega^2 \varphi=f. \end{align*}

Solution: 1) Let $x\in\mathbb{R}$ and $t\in\mathbb{R}$. We know from trigonometric formulas that \begin{align*} \sin(\omega x-\omega t)=\sin(\omega x)\cos(\omega t)-\cos(\omega x)\sin(\omega t). \end{align*} Then \begin{align*} \varphi(x)&= \frac{\sin(\omega x)}{\omega} \int^x_0 \cos(\omega t)f(t)dt-\frac{\cos(\omega x)}{\omega} \int^x_0 \sin(omega t)f(t)dt. \end{align*} Thus it suffices to select \begin{align*} \Phi(x)= \int^x_0 \cos(\omega t)f(t)dt\quad\text{and} \quad \Psi(x)=\int^x_0 \sin(\omega t)f(t)dt. \end{align*}

2) The functions $\Phi$ and $\Psi$ are primitives of continuous functions. Thus $\Phi$ and $\Psi$ are $C^1$ functions on $\mathbb{R}$ and that \begin{align*} \Phi'(x)=\cos(\omega x)f(x),\qquad \Psi(x)=\sin(\omega x)f(x) \end{align*} for any $x\in \mathbb{R}$. This proves that the function $varphi$ is differential on $\mathbb{R}$ as the product and sum of differentiable functions. Moreover, \begin{align*} \omega\varphi'(x)&= \omega \cos(\omega x) \Phi(x)+\sin(\omega x) \cos(\omega x)f(x)\cr & \hspace{2cm}+\omega\sin(\omega x)\Psi(x)-\cos(\omega x)\sin(\omega x)f(x)\cr &= \omega \cos(\omega x) \Phi(x)+\omega\sin(\omega x)\Psi(x). \end{align*} As $\omega\neq 0,$ then \begin{align*} \varphi'(x)=\cos(\omega x) \Phi(x)+\sin(\omega x)\Psi(x) \end{align*} This last formula show that $\varphi’$ is differentiable as product and sum of differentiable functions. Hence $\varphi$ is twice differentiable and \begin{align*} \varphi”(x)&=-\omega\sin(\omega x) \Phi(x)+\cos^2(\omega x)f(x)\cr & \hspace{2.5cm}+\omega\cos(\omega x)\Psi(x)+\sin^2(\omega x)f(x)\cr &= -\omega (\sin(\omega x) \Phi(x)-\cos(\omega x)\Psi(x))+f(x)\cr &= -\omega^2 \varphi(x)+f(x) \end{align*} for any $x\in\mathbb{R}$. This proves that \begin{align*} \varphi”+\omega^2 \varphi=f. \end{align*}

We discuss the properties of the translation operator defined in a Lebesgue space. The latter appears in the study of differential equations, mainly equations with negative memories called also delay equations.

The semigroup property of the translation operator 

Let $p\ge 1$ be a real number and $E=L^p([0,+\infty)$ be the space of all measurable functions such that \begin{align*} \int^{+\infty}_0 |f(s)|^pds <\infty.\end{align*} If we select \begin{align*} \|f\|_p:=\left( \int^{+\infty}_0 |f(s)|^pds <\infty \right)^{\frac{1}{p}}\end{align*}\then $(E,\|\cdot\|)$ is a Banach space of infinite dimension, the Lebesgue space. On the other hand, let us define \begin{align*} (T_t f)(s)=f(t+s),\qquad \forall s,t\ge 0,\; f\in E.\end{align*}

Fact 1: for any $t\ge 0,$ $T_t$ is a linear continuous operator on $E,$ i.e. $T_t\in \mathscr{L}(E)$. Moreover, $\|T_t\|_{\mathscr{L}(E)}\le 1$ for any $t$.

Proof: Let $t\ge 0$ and $f\in E$. The function $s\mapsto (T_t f)(s)=f(t+s)$ is measurable because as composition of two measurables functions, $f$ and the function $s\mapsto t+s$. Moreover, we estimate \begin{align*} \int^{+\infty}_0 |(T_tf)(s)|^pds&= \int^{+\infty}_0 |f(t+s)|^pds \cr & =\int_t^{+\infty}|f(\sigma)|^pd\sigma\cr & \le \int^{+\infty}_0 |f(\sigma)|^pd\sigma=\|f\|_p^p.\end{align*} This shows that $T_f f\in E$, and \begin{align*} \|T_t f\|_p\le \|f\|_p,\qquad \forall t\ge 0,\;f\in E.\end{align*}

Definition: The operator $T_t$ is called the translation operator or the shift operator.

Fact 2: $T_0=Id_E$ and for any $t,s\ge 0,$ $T_{t+s}=T_tT_s$, here $T_tT_s:=T_t\circ T_s$.

Proof: For $f\in E$ and $t,s\ge 0,$ we have \begin{align*} (T_{t+s}f)(s)=f(s+(t+\sigma))=(T_sf)(t+\sigma)=(T_t T_sf)(\sigma).\end{align*}Definition: The family of operators $(T_t)_{t\ge 0}$ satisfying the fact 1 and the fact 2 is called the semigroup of operators or the shift semigroup.

The strong continuity of the shift semigroup 

In this section, we show that the translation operator $T_t$ satisfies a nice topological property.

Fract 3: The family $(T_t)_{t\ge 0}$ is strongly continuous at zero. That is for any $f\in E,$ we have $$\lim_{t\to 0}\|T_t f-f\|_p=0.$$

Proof:  For $f\in E,$ we select $u(t):=T_tf$. According to Fact 2, it suffices to prove that $\|u(t)-u(0)\|_p\to 0$ as $t\to 0$. Before solving this question, we first need to recall some facts. First, a function $f$ has a support compact if it vanishes outside a compact. The support of $f$ is the compact set \begin{align*}{\rm supp}(f):=\overline{\{t: f(t)\neq 0\}}.\end{align*} Second, we recall that the set of continuous functions with compact support, $\mathscr{C}_c([0,+\infty))$, is dense in $L^p([0,+\infty))$.

As $f\in E,$ then by density there exists a sequence $(f_n)_n\subset \mathcal{C}_c([0,+\infty)) $ such that $\|f_n-f\|_p\to 0$ as $n\to+\infty$. For any $\varepsilon,$ there exists $m\in \mathbb{N}$ such that $\|f_m-f\|\le \frac{\varepsilon}{3}$. On the other hand, we have\begin{align*}\|u(t)-u(0)\|&=\|T_t f-f\|_p\cr & \le \|T_t f-T_t f_m\|_p + \|T_t f_m-f_m\|_p+\|f_m-f\|_p\cr & \le 2 \|f_m-f\|_p + \|T_t f_m-f_m\|_p \cr &\le \frac{2}{3}\varepsilon+ \|T_t f_m-f_m\|_p .\end{align*}

Now it suffices to show that $\|T_t f_m-f_m\|_p\to 0$ as $t\to 0$. In fact, put $K={\rm supp}(f)$. Then, \begin{align*} \|T_t f_m-f_m\|_p ^p&=\int_K |f_m(t+s)-f(s)|^p ds\cr &\le {\rm meas}(K) \left(\sup_{s\in K}|f(t+s)-f(s)|\right )^p .\end{align*} Here $ {\rm meas}(K) $ is the Lebesgue measure of the set $K$. As the function $f$ is uniformly continuous on the compact set, then we have the following uniform convergence \begin{align*}\lim_{t\to 0} \sup_{s\in K}|f(t+s)-f(s)| =0.\end{align*} Then there exists $\delta>0$ such that for any $t\in (0,\delta)$, we have \begin{align*} \|T_t f_m-f_m\|_p \le \frac{\varepsilon}{3}.\end{align*}Thus, for any $\varepsilon,$ there exists $\delta>0$ such that for any $t,$ \begin{align*} t\in (0,\delta)\Longrightarrow \|u(t)-u(0)\| < \varepsilon.\end{align*}

Primitives of continuous functions play a key role in the calculation of integrals. Therefore, in this article, we will use examples to introduce a simple technique for determining continuous function primitives.

We assume that the reader knows how to calculate integrals.

Generalities on primitives of continuous functions

Let $f:D\subset \mathbb{R}$ be a continuous function. The primitive of the function $f$ is a differentiable function $F:D\to\mathbb{R}$ such that $F'(x)=f(x)$ for any $x\in D$.

If $f$ admits a primitive $F$, then the function $G=F+c$ is a primitive of $f$ for any constant $c\in\mathbb{R}$. Two primitives of $f$ differ from a constant.

Let us now state the fundamental theorem of calculus: Let $f:D\to \mathbb{R}$ be a continuous function. The n for any $a\in D,$ the function defined by We also write \begin{align*}F(x)=\int^x_a f(t)dt,\quad x\in D,\end{align*} is differentiable on $D$ and $F’=f$.

Remark that the primitive a function of $C^1$ class. This means that the derivative function $F’$ is continuous.

Examples of how to find primitives 

To determine a primitive of a continuous function, you need to know some classical primitives as well as partial fraction decomposition rules.

• Determine a primitive of the function \begin{align*} f(x)=\frac{2x^2-3x+4}{(x-1)^2}\textrm{ on }]1,+\infty[ .\end{align*} We look for real constants $a,b$ and $c$ such that \begin{align*} f(x)=a+\frac{b}{x-1}+\frac {c}{(x-1)^2}. \end{align*} By putting everything at the same denominator in the right-hand side, we find \begin{align*} f(x)=\frac{ax^2+x(-2a+b)+(a-b+c)}{(x-1)^2}. \end{align*} According to the first expression of $f,$ one obtains $a=2,;b=1$ and $c=3$, so that for any $x>1,$ \begin{align*} f(x)=2+\frac{1}{x-1}+\frac {3}{(x-1)^2}. \end{align*} The primitive of $f$ on $]1,+\infty[$ is then given by \begin{align*} F(x)=2x+\ln(x-1)-\frac {3}{x-1}+C,\qquad C\in \mathbb{R} \end{align*}
• Determine the primitive of the function \begin{align*} f(x)=\frac{2x-1}{1+x^2}\textrm{ on }\mathbb{R}.\end{align*} We first observe that $f$ is defined on $\mathbb{R}$. On the other hand, we can write \begin{align*} f(x)&=\frac{2x}{1+x^2}-\frac{1}{1+x^2}\cr &=\frac{(1+x^2)’}{1+x^2}-\frac{1}{1+x^2}\cr &= \frac{d}{dx}\left(\ln(1+x^2)-\arctan(x)\right). \end{align*} Thus, the primitive of $f$ on $\mathbb{R}$ is given by \begin{align*} F(x)=\ln(1+x^2)-\arctan(x)+C,\qquad C\in\mathbb{R}. \end{align*}
• Calculate the primitive of the following function\begin{align*} f(x)=x e^{\lambda x}\textrm{ on }\mathbb{R}.\end{align*} In this question, we will use integration by parts method. We then have for any $\lambda\neq 0,$ \begin{align*} \int^x_c t e^{\lambda t}dt&= \int^x_c \frac{t}{\lambda} \left(e^{\lambda t}\right)’dt\cr & =\left[\frac{t}{\lambda} e^{\lambda t}\right]^x_c-\int^x_c \frac{1}{\lambda}e^{\lambda t}dt. \end{align*} Hence the primitive of $f$ on $\mathbb{R}$ is given by \begin{align*} F(x)=\frac{1}{\lambda} xe^{\lambda x}-\frac{1}{\lambda^2} e^{\lambda x}+C,\qquad C\in\mathbb{R}. \end{align*}
• Compute the primitive of the function\begin{align*} f(x)=\frac{1}{\sqrt{x^2+2x+5}}\textrm{ on }\mathbb{R}.\end{align*} Observe that $f$ is defined on all $\mathbb{R}$. Here we shall use the change of variables technique. We have \begin{align*} \int^x_c \frac{dt}{\sqrt{t^2+2t+5}}&=\int^x_c \frac{dt}{\sqrt{(t^2+2t+4)+1}}\cr &=\int^x_c \frac{dt}{\sqrt{(t+2)^2+1}}. \end{align*} We now put $u=t+2$ then $du=dt$ and \begin{align*} \int^x_c \frac{dt}{\sqrt{t^2+2t+5}}&=\int^{x+2}_{c+2} \frac{du}{\sqrt{u^2+1}}\cr &= 2\left[\ln(u+\sqrt{u^2+1})\right]^{x+2}_{c+2}\cr &= 2\ln(x+2+\sqrt{x^2+2x+5})+C. \end{align*} This implies that the primitive of $f$ is \begin{align*} F(x)=2\ln(x+2+\sqrt{x^2+2x+5})+C. \end{align*}